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# Light of wavelength $5000\overset{{}^\circ }{\mathop{\text{A}}}\,$ falls on a sensitive plate with photoelectric work function of $1.9eV$ . The maximum kinetic energy of the photoelectron emitted will be.A. $1.16eV$B. $2.38eV$C. $0.58eV$D. $2.98eV$

Last updated date: 13th Aug 2024
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Hint: First of all we will find the incident energy of light by using the formula, ${{E}_{i}}=\dfrac{hc}{\lambda }$ and then we will convert it into electron volt i.e. ${{E}_{v}}$. Then, we will find the maximum kinetic energy of the proton emitted by taking the difference of energy of electron volt and photoelectric function.

Formula used: ${{E}_{i}}=\dfrac{hc}{\lambda }$

In the question we are given that light of wavelength $5000\overset{{}^\circ }{\mathop{\text{A}}}\,$ falls on a sensitive plate with photoelectric work function of $1.9eV$, so first of all we will find the energy of the light by using the formula,
${{E}_{i}}=\dfrac{hc}{\lambda }$ …………………(i)
Where, $\lambda$ is the wavelength of the light, c is velocity if light and h is the plank constant.
Now, in question it is given that wavelength of light is $5000\overset{{}^\circ }{\mathop{\text{A}}}\,$and in order to find the incident energy in terms of electron volt the value of $hc$ can be given as 12375. So, on substituting the values in equation (i) we will get,
${{E}_{i\left( eV \right)}}=\dfrac{hc}{\lambda }=\dfrac{12375}{5000\times {{10}^{-10}}}=2.475eV$
Now, we will find the maximum kinetic energy emitted by the proton by taking the difference of electron energy and proton energy as,
${{\alpha }_{\max }}={{E}_{i\left( eV \right)}}-\varphi$
${{\alpha }_{\max }}=2.475-1.9=0.575eV\cong 0.58eV$.
Thus, maximum kinetic energy can be given as $0.58eV$.
Hence, option (a) is the correct answer.

Note: Students might forget to convert the incident energy into electron volt energy and due to that they might not get the desired answer. Students should also know the values of Planck’s constant and velocity of light to convert it into electron volt energy.